# Molecular Hamiltonian

I was reading some material on the Molecular Hamiltonian on Wiki. It said that,

Almost all calculations of molecular wavefunctions are based on the separation of the Coulomb Hamiltonian first devised by Born and Oppenheimer. The nuclear kinetic energy terms are omitted from the Coulomb Hamiltonian and one considers the remaining Hamiltonian as a Hamiltonian of electrons only.

I am not sure why this is done. This won't be the "full" Hamiltonian if we omit the KE terms of the nuclei. Can somebody please explain to me why the nuclear kinetic energy terms can be ignored from the Coulomb Hamiltonian.

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As far as I can remember my (not so fortunate) atomic & molecular physics class, the idea is to calculate the time and length scales of both systems. If you try to estimate it for typical situations you find you that they are very different, and so, 'one dynamics should not mix with the other', in the following sense: The electrons would be much faster than the nuclei, so they see the nuclei as a static (and so, drop the kinectic term), and the nuclei only see an average of the electron's movement, but not it full dynamics. –  Hydro Guy May 29 '13 at 3:21

It is important to indicate that the Hamiltonian is not decoupled: it is strongly coupled to the position of the nuclei, but not to the momenta. Thus in the BO approximation the full wave function implies strong correlation of electronic and nuclear motion: $\phi^{nuc}(\bf{R})\psi^{el}(\bf{r};\bf{R})$. If the coupling could be neglected the wave function would be separable: $\phi^{nuc}(\bf{R})\psi^{el}(\bf{r})$. While this is a poor approximation, BO gives excellent results except when the kinetic energy of the nuclei is of the order of the energy difference between adjacent electronic states. –  perplexity Aug 27 '13 at 9:42